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Author Question: Solve the initial-value problem y" + 2 = 0, y'(2) = 3, y(-1) = -4. (Read 99 times)

biggirl4568

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on: May 27, 2021

Question 1

Solve the integral equation  y(x) = 1 + dt  by reinterpreting it as an initial-value problem.
◦ y =
◦ y =
◦ y =
◦ y =
◦ y =

Question 2

Solve the initial-value problem  y" + 2 = 0,   y'(2) = 3,   y(-1) = -4.
◦ y = -x2 - x - 4
◦ y = -x2 + 7x + 4
◦ y = -x2 + 7x + 12
◦ y = -x2 + 7x - 12
◦ y = -x2 + x - 2
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Marked as best answer by biggirl4568 on May 27, 2021

hanadaa

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Reply #1 on: May 27, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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biggirl4568

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Reply 2 on: May 27, 2021
Gracias!

cassie_ragen

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Reply 3 on: Yesterday
Excellent

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